inequality - $left(1+frac{1}{n^2-1}right)^n le n^{frac{1}{n}}$ for sufficiently large $n$

Is the following true $\left(1+\frac{1}{n^2-1}\right)^n \le n^{\frac{1}{n}}$ for sufficiently large $n$? I'm sure it is, but I'm having difficulty proving it. I've tried using the Bernoulli inequality and Arithmetic-Geometric mean inequality but to no avail. Is there an elegant way of doing this?

Answer

$$\begin{align}
\left(1+\frac{1}{n^2-1}\right)^{n^2-1}&\left(1+\frac{1}{n^2-1}\right)^{n^2}&\end{align}$$
for $n\geq6$. Then raise both sides to the $1/n$. Then check $n=2,3,4,5$ directly. (By the way, it's false for $n=2$.)

But it's actually a pretty bad bound. This argument shows the much tighter $$\left(1+\frac{1}{n^2-1}\right)^{n}<(2e)^{1/n}$$